Cluster Points and Sequence Limits
A sequence can return to a neighborhood infinitely often without eventually staying there. A point with that repeated-neighborhood property is a cluster point of the sequence. The key phrase is “infinitely many terms”: terms are counted by their indices, so repeated occurrences of the same value count separately. This makes cluster points different from limits, which describe what happens to all sufficiently late terms.
Let \((X,\mathcal{T})\) be a topological space, and let \((x_n)\) be a sequence in \(X\). For an open neighborhood \(U\) of \(x\), record the indices at which the sequence lies in \(U\): \(\{n\in\mathbb{N}:x_n\in U\}\). The set of indices, rather than the set of distinct values, is what we count.
Thus a limit requires eventual membership in each neighborhood, while a cluster point requires membership at infinitely many indices. A sequence may have cluster points that are not limits, and it may have no limit even when it has cluster points.
A Tail Test for Cluster Points
Proof. Suppose \(x\) is a cluster point. Fix an open neighborhood \(U\) of \(x\) and an index \(N\). If no \(n\ge N\) satisfied \(x_n\in U\), then all indices at which the sequence lies in \(U\) would belong to the finite set \(\{1,\ldots,N-1\}\). This contradicts the definition of a cluster point. Conversely, suppose the stated tail condition holds. If the set of indices with \(x_n\in U\) were finite, it would have an upper bound \(M\); taking \(N=M+1\) would contradict the tail condition. If that set is empty, taking \(N=1\) gives the same contradiction. Therefore every neighborhood contains terms at infinitely many indices, and \(x\) is a cluster point. \(\square\)
This test says that no matter how far along the sequence we start, some later term must return to each neighborhood of a cluster point. It also shows why changing finitely many terms cannot change the cluster points: after any finite change, the tail test has the same outcome.
Proof. Suppose \(x_n\to x\). Let \(U\) be any open neighborhood of \(x\). By convergence, there is an \(N\) such that \(x_n\in U\) for every \(n\ge N\). The infinitely many indices \(N,N+1,N+2,\ldots\) therefore all have their terms in \(U\). This holds for every such \(U\), so \(x\) is a cluster point. \(\square\)
The reverse implication does not hold in general. A cluster point guarantees infinitely many visits, not that the sequence eventually stays nearby.
Worked Example: An Alternating Sequence in the Real Line
Give \(\mathbb{R}\) its standard topology and define \(x_{2k-1}=1\), \(x_{2k}=-1\) for \(k\in\mathbb{N}\). Find all cluster points and determine whether the sequence has a limit.
Every open neighborhood of \(1\) contains \(1\), and \(x_{2k-1}=1\) for every \(k\). It therefore contains terms at infinitely many odd indices, so \(1\) is a cluster point. Every open neighborhood of \(-1\) contains terms \(x_{2k}=-1\) at infinitely many even indices, so \(-1\) is also a cluster point.
If \(y\in\mathbb{R}\) is different from \(1\) and \(-1\), let \(r=\tfrac12\min\{|y-1|,|y+1|\}\). Both distances in the minimum are positive, so \(r>0\). The open interval \((y-r,y+r)\) contains neither \(1\) nor \(-1\). Since every term of the sequence is one of those two values, this neighborhood contains no terms. Thus \(y\) is not a cluster point, and the cluster-point set is exactly \(\{-1,1\}\).
The sequence does not converge. If it converged to some \(y\ne 1\), the open interval centered at \(y\) with radius \(|y-1|/2\) would exclude \(1\), but infinitely many odd-indexed terms equal \(1\); hence the sequence could not eventually lie in that interval. If it converged to \(1\), the neighborhood \((1/2,3/2)\) would have to contain all sufficiently late terms, but infinitely many even-indexed terms equal \(-1\), which is outside the interval. Both possibilities fail.
Check: the neighborhood \((1/2,3/2)\) contains \(1\) at all odd indices and contains no \(-1\) terms. The neighborhood \((-3/2,-1/2)\) contains \(-1\) at all even indices and contains no \(1\) terms. These counts confirm the two cluster points and the failure of eventual membership required for convergence.
Repeated Terms and the Meaning of “Infinitely Many”
When a value occurs repeatedly, every occurrence contributes a separate index. Conversely, a value that occurs only once does not by itself make a point a cluster point: every neighborhood of a cluster point must contain terms at infinitely many indices. The next example also illustrates that a cluster point need not be a value appearing in the sequence.
Worked Example: A Repeated Value and Single Occurrences
In \(\mathbb{R}\) with its standard topology, define \(x_{3k-2}=0\), \(x_{3k-1}=0\), and \(x_{3k}=k\) for \(k\in\mathbb{N}\). The sequence begins \(0,0,1,0,0,2,0,0,3,\ldots\). Find all cluster points and determine whether it converges.
Every open neighborhood of \(0\) contains the terms at indices \(3k-2\) and \(3k-1\) for every \(k\). There are infinitely many such indices, so \(0\) is a cluster point.
Now let \(y\ne0\). Choose \(r=|y|/2>0\). The neighborhood \(U=(y-r,y+r)\) does not contain \(0\): if \(y>0\), its left endpoint is \(y/2>0\), and if \(y<0\), its right endpoint is \(y/2<0\). The only terms that can lie in \(U\) are the values \(k\in\mathbb{N}\). The interval \(U\) is bounded, so there are only finitely many positive integers in it. Consequently \(U\) contains terms at only finitely many indices, and \(y\) is not a cluster point. The full cluster-point set is \(\{0\}\).
The sequence does not converge. If it converged to \(y\), it would eventually lie in some bounded open interval around \(y\). But the terms \(x_{3k}=k\) are unbounded: for any upper endpoint \(b\), choose \(k>b\), and then \(x_{3k}=k>b\). In fact, arbitrarily late such terms lie outside the interval. Thus no \(y\) can be a limit.
Check: \(0\) occurs at the two indices \(3k-2\) and \(3k-1\) for every \(k\), giving infinitely many occurrences. Each positive integer \(k\) occurs only at index \(3k\); any bounded neighborhood contains only finitely many such integer values. This verifies the index counts used above.
Worked Example: A Repeated Value in a Convergent Sequence
In \(\mathbb{R}\) with its standard topology, define \(x_{2k-1}=0\) and \(x_{2k}=1/k\) for \(k\in\mathbb{N}\). Find all cluster points and all limits.
Every neighborhood of \(0\) contains all the odd-indexed terms, which equal \(0\), so \(0\) is a cluster point. The sequence also converges to \(0\). To verify this, let \(\varepsilon>0\) and choose \(K\in\mathbb{N}\) with \(K>1/\varepsilon\). For \(k\ge K\), \(1/k\le1/K<\varepsilon\), and the odd-indexed terms are exactly \(0\). Hence all terms from index \(2K-1\) onward have distance less than \(\varepsilon\) from \(0\).
No \(y\ne0\) is a cluster point. Choose \(r=|y|/2\), so the neighborhood \((y-r,y+r)\) excludes \(0\). If \(y<0\), it also excludes every \(1/k>0\), and contains no terms. If \(y>0\), any term \(1/k\) in this interval must satisfy \(1/k>y/2\), hence \(k<2/y\). Only finitely many positive integers \(k\) satisfy this inequality, so the neighborhood contains terms at only finitely many indices. Thus \(0\) is the only cluster point. Since every limit is a cluster point, the only possible limit is \(0\); the convergence calculation above confirms that it is a limit.
Check: the odd-indexed terms give infinitely many occurrences of \(0\). For the even-indexed terms, \(k\ge K>1/\varepsilon\) gives \(1/k\le1/K<\varepsilon\), so both types of terms are eventually in \((-\varepsilon,\varepsilon)\), as required for convergence.
Cluster Points and Convergent Subsequences
A subsequence is obtained by selecting terms at strictly increasing indices \(n_1<n_2<\cdots\). In a first-countable space, cluster points can also be described as limits of subsequences. A space is first countable if every point has a countable local base: a countable collection of open neighborhoods such that every open neighborhood of the point contains a member of that collection.
Proof. First suppose \(x\) is a cluster point, and choose a countable local base \(B_1,B_2,\ldots\) at \(x\). Define \(W_j=B_1\cap\cdots\cap B_j\). Each \(W_j\) is an open neighborhood of \(x\), since it is a finite intersection of open neighborhoods. Given any open neighborhood \(U\) of \(x\), some \(B_i\) satisfies \(B_i\subseteq U\); for every \(j\ge i\), this gives \(W_j\subseteq B_i\subseteq U\).
Choose indices recursively. Since \(x\) is a cluster point, every neighborhood \(W_j\) contains terms at infinitely many indices. We can therefore choose \(n_j>n_{j-1}\) with \(x_{n_j}\in W_j\), starting with any \(n_1\) for which \(x_{n_1}\in W_1\). For any neighborhood \(U\) of \(x\), choose \(i\) with \(B_i\subseteq U\). For all \(j\ge i\), \(x_{n_j}\in W_j\subseteq U\). Thus \(x_{n_j}\to x\).
Conversely, suppose a subsequence \((x_{n_j})\) converges to \(x\). For any open neighborhood \(U\) of \(x\), all sufficiently late terms of the subsequence lie in \(U\). Their indices \(n_j\) are strictly increasing, so this gives infinitely many distinct indices of the original sequence whose terms lie in \(U\). Hence \(x\) is a cluster point. \(\square\)
The countable local base matters in the forward direction: it provides a countable sequence of neighborhoods whose tests ensure convergence. The cluster-point definition itself needs no countability assumption. In \(\mathbb{R}\) with its standard topology, intervals centered at \(x\) with radii \(1/j\) form a countable local base, so the theorem applies.
A Common Counting Pitfall
Common mistake: “A neighborhood must contain infinitely many different values to make its point a cluster point.” That incorrectly discards repeated terms. For the constant sequence \(x_n=0\) in \(\mathbb{R}\), the neighborhood \((-1,1)\) contains only one distinct value from the sequence, namely \(0\). But it contains \(x_n\) at every index \(n\in\mathbb{N}\), so \(0\) is a cluster point. It is also a limit, since every neighborhood of \(0\) contains all terms. The definition counts indices, not distinct values.
Keep the two tests separate: a limit requires that every neighborhood contain all terms after some index; a cluster point requires that every neighborhood contain terms at infinitely many indices. Every limit passes the second test, but the alternating sequence shows that passing the second test does not guarantee the first.
Check Your Understanding
For each question, count indices rather than distinct values when applying the cluster-point definition.
- In \(\mathbb{R}\), find the cluster points and limits of \(x_n=(-1)^n\).
- For the sequence \(0,0,1,0,0,2,0,0,3,\ldots\), explain why \(0\) is a cluster point and why \(2\) is not.
- Can a cluster point of a sequence fail to be one of its term values? Give an example and verify the neighborhood test.
- Prove directly from the definitions that changing finitely many terms of a sequence does not change its cluster points.
- Explain why the tail test for cluster points requires a term in every neighborhood after every starting index.